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Thread: True or false analysis.

  1. #1
    Super Member Showcase_22's Avatar
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    True or false analysis.

    Which of the following are true and which false? Give proofs or counterexamples.
    i). If $\displaystyle x \rightarrow f(x)^2$ (this is a mapping, I couldn't quite find the right arrow) is continous at c then f is differentiable at c.
    I said this was false and my counterexample was $\displaystyle f(x)=| \sqrt{x}|$.

    This function is continous at 0 but $\displaystyle f(x)^2=|x|$ which is not differentiable at 0.

    ii). If $\displaystyle \exists \ ga,b) \rightarrow \mathbb{R}$ and $\displaystyle Za,b) \rightarrow \mathbb{R}$ such that $\displaystyle f(x)=g(x)Z(x)$ then f is differentiable at c if:

    a). Z is bounded
    b). g is differentiable at c with g'(c)=0 and
    c). g(c)=0;
    I'm not really sure about this one.

    I think it's true but I don't know how to prove it.

    iii). In ii) f is differentiable assuming only (a) and (c).
    I put false and used the counterexample $\displaystyle f(x)=x \sin \left( \frac{1}{x} \right)$ when c=0.

    $\displaystyle Z(x)= \sin \left( \frac{1}{x} \right)$ and $\displaystyle g(x)=x$.

    In ii). f is differentiable assuming only (a) and (b).
    Once again I put false and used the counterexample $\displaystyle f(x)=5 \sin \left( \frac{1}{x} \right)$ where c=0.

    Where $\displaystyle g(x)=5$ and $\displaystyle Z(x)= \sin \left( \frac{1}{x} \right)$

    In (ii). f is differentiable assuming only (b) and (c).
    This is false because $\displaystyle Z(x)$ could be discontinous.

    My counterexample was $\displaystyle f(x)=x^2. \begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\
    0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases}$ at c=0.

    Here $\displaystyle g(x)=x^2$ and $\displaystyle Z(x)=\begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\
    0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases}$.

    are these correct?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    I said this was false and my counterexample was $\displaystyle f(x)=| \sqrt{x}|$.

    This function is continous at 0 but $\displaystyle f(x)^2=|x|$ which is not differentiable at 0.
    You mean $\displaystyle f(x)=\sqrt{|x|}$ I suppose ?
    If so it is OK
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    I'm not really sure about this one.

    I think it's true but I don't know how to prove it.
    Yes it is true

    $\displaystyle \frac{f(x) - f(c)}{x-c} = \frac{g(x)Z(x) - g(c)Z(c)}{x-c} = \frac{g(x)Z(x)}{x-c} = Z(x) \:\frac{g(x)}{x-c} = Z(x) \:\frac{g(x) - g(c)}{x-c}$

    I think that you can finish.
    This will give you ideas for the other cases.
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    I put false and used the counterexample $\displaystyle f(x)=x \sin \left( \frac{1}{x} \right)$ when c=0.

    $\displaystyle Z(x)= \sin \left( \frac{1}{x} \right)$ and $\displaystyle g(x)=x$.
    I do not agree with your counter-example since g is not defined in 0 (even if its limit is 0).

    Take $\displaystyle Z(x) = 1 \mbox{ and } g(x) = \sqrt{|x|}$ over (-1,1) for instance

    Z is bounded and g(0)=0 therefore (a) and (c) are verified

    $\displaystyle f(x) = \sqrt{|x|}$ is not differentiable in 0
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